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7.Gravitation
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The mean radius of the earth's orbit round the sun is $1.5 \times {10^{11}}$. The mean radius of the orbit of mercury round the sun is $6 \times {10^{10}}\,m$. The mercury will rotate around the sun in
A
A year
B
Nearly $4$ years
C
Nearly $\frac{1}{4}$ year
D
$2.5$ years
Solution
(c) $\frac{{{T_{{\rm{mercury}}}}}}{{{T_{{\rm{earth}}}}}} = {\left( {\frac{{{r_{{\rm{mercury}}}}}}{{{r_{{\rm{earth}}}}}}} \right)^{3/2}} = {\left( {\frac{{6 \times {{10}^{10}}}}{{1.5 \times {{10}^{11}}}}} \right)^{3/2}} = \frac{1}{4}$ (approx.)
$\therefore$ ${T_{{\rm{mercury}}}} = \frac{1}{4}{\rm{ year}}$
Standard 11
Physics
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